Lecture 4 Engineering Economics a detail lecutrer.pptx

Chapter 2
Factors: How Time
and Interest Affect
Money
Lecture slides to accompany
Engineering Economy
8th edition
Leland Blank
Anthony Tarquin
© 2012 by McGraw-Hill, New York, N.Y All RightsReserved
2-1

LEARNING OUTCOMES
1. F/P and P/F Factors
2. P/A and A/P Factors
3. F/A and A/F Factors
4. Factor Values
5. Arithmetic
Gradient 6.
Geometric Gradient
7. Find i or n
2-2

Timing of Cash Flows and Modelling
• The actual timing of cash flows can be very complicated and irregular. Unless
some simple approximation is used, comparisons of different cash flow
sequences will be very difficult and impractical.
• Consider, for example, the cash flows generated by a relatively simple
operation like a service station that sells gasoline, supplies and also services
cars.
• Some cash flows, like sales of gasoline and minor supplies, will be almost
continuous during the time the station is open.
• Other flows, like receipts for the servicing of cars, will be on a daily basis.
• Disbursements for wages may be on a weekly basis. Some disbursements,
like those for a manager’s salary and for purchases of gasoline and supplies,
may be monthly. Disbursements for insurance and taxes may be quarterly or
semiannually. Other receipts and disbursements, like receipts for major
repairs or disbursements for used parts, may be irregular.

Timing of Cash Flows and Modelling
• Because of the difficulties of making precise calculations of complex and irregular
cash flows, engineers usually work with fairly simple models of cash flow patterns.
• The most common type of model assumes that all cash flows and all compounding of
cash flows occur at the ends of conventionally defined periods, such as months or
years. Models that make this assumption are called discrete models.
• In some cases, analysts use models that assume cash flows and their compounding
occur continuously over time; such models are called continuous models.

Compound Interest Factors for Discrete
Compounding
Compound interest factors are formulas that define mathematical equivalence for
specific common cash flow patterns.
The four patterns are:
1. A single disbursement (money spent) or receipt (money received)
2. A set of equal disbursements or receipts over a sequence of periods, referred to as
an annuity
3. A set of disbursements or receipts that change by a constant amount from one
period to the next in a sequence of periods, referred to as an arithmetic gradient series
4. A set of disbursements or receipts that change by a constant proportion from one
period to the next in a sequence of periods, referred to as a geometric gradient series

Compound Interest Factors for Discrete
Compounding
The principle of discrete compounding requires several
assumptions:
1. Compounding periods are of equal length.
2. Each disbursement and receipt occurs at the end of a period.

Compound Interest Factors for Single
Disbursements or Receipts
• In many situations, a single disbursement or receipt is an
appropriate model of cash flows.
• For example, the salvage value of production equipment
with a limited service life will be a single receipt at some
future date.
• An investment today to be redeemed at some future date is
another example.

Compound Interest Factors for Single
Disbursements or Receipts
• The compound amount factor, denoted by
(F/P,i,n), gives the future amount, F, that is
equivalent to a present amount, P, when the
interest rate is i and the number of periods is n.

Single Payment Factors (F/P and P/F)
Single payment factors involve only P and F. Cash flow diagrams are as follows:
F = P(1 + i ) n P = F[1 / (1 + i ) n]
Formulas are as follows:
Terms in parentheses or brackets are called factors. Values are in tables for i and n values
Factors are represented in standard factor notation such as (F/P,i,n),
where letter to left of slash is what is sought; letter to right represents what is given
2-9

The compound amount factor and the present worth factor
are fundamental to engineering economic analysis.
Their most basic use is to convert a single cash flow that
occurs at one point in time to an equivalent cash flow at
another point in time.

F/P and P/F for Spreadsheets
2-11
Future value F is calculated using FV function:
= FV(i%,n,,P)
Present value P is calculated using PV function:
= PV(i%,n,,F)
Note the use of double commas in each function

Example: Finding Future Value
A person deposits $5000 into an account which pays interest at a rate of 8%
per year. The amount in the account after 10 years is closest to:
(A) $2,792 (B) $9,000 (C) $10,795 (D) $12,165
The cash flow diagram is:
Answer is (C)
2-12
Solution:
F = P(F/P,i,n )
= 5000(F/P,8%,10 )
= 5000(2.1589)
= $10,794.50

Example: Finding Present Value
A small company wants to make a single deposit now so it will have enough money to
purchase a backhoe costing $50,000 five years from now. If the account will earn
interest of 10% per year, the amount that must be deposited now is nearest to:
(A) $10,000 (B) $ 31,050 (C) $ 33,250 (D) $319,160
The cash flow diagram is: Solution:
P = F(P/F,i,n )
= 50,000(P/F,10%,5 )
= 50,000(0.6209)
= $31,045
Answer is (B)
2-13

Example 2.1
Sandy, a manufacturing engineer, just received a
year-end bonus of $10,000 that will be invested
immediately. With the expectation of earning at the
rate of 8% per year, Sandy hopes to take the entire
amount out in exactly 20 years to pay for a family
vacation when the oldest daughter is due to graduate
from college. Find the amount of funds that will be
available in 20 years by using (a) hand solution by
applying the factor formula and tabulated value, and
(b) a spread sheet function.

Lecture 4 Engineering Economics a detail lecutrer.pptx

Example 2.2
The Steel Plant Case: Hardee Steel, a subsidiary of the
international consortium Beleez Group, headquartered in Italy,
has announced plans to develop a state-of-the art steel and
customized products fabrication plant in North America. The
selected location is ideal because there is a large deposit of iron
ore in the area. The new company will be named Hardee Beleez,
NA, Inc. or HBNA. The plant investment, expected to amount to
$200 million, is scheduled for 2020. When the plant is
completed and operating at full capacity, based upon the
projected needs and cost per metric ton, it is possible that the
plant could generate as much as $50,000,000 annually in
revenue. All analyses will use a planning horizon of 5 years
commencing when the plant begins operation.

Example 2.2
The HBNA plant will require an investment of $200 million to
construct. Delays beyond the anticipated implementation year
of 2020 will require additional money to construct the plant.
Assuming that the cost of money is 10% per year compound
interest, use both tabulated factor values and spreadsheet
functions to determine the following for the board of directors
of the Italian company that plans to develop the plant.
(a) The equivalent investment needed in 2023 if the plant is
delayed for 3 years.
(b) The equivalent investment needed in 2023 if the plant is
constructed sooner than originally planned.

F = 1,200,000(F/P,7%,4)
= 1,200,000(1.3108)
= $1,572,960

P = 7(120,000)(P/F,10%,2)
= 840,000(0.8264)
= $694,176

P = 30,000,000(P/F,10%,5) –
15,000,000
= 30,000,000(0.6209) – 15,000,000
= $3,627,000

Compound Interest Factors for Annuities
A series of uniform receipts or disbursements that start at
the end of the first period and continue over N periods.
Mortgage or lease payments and maintenance contract fees
are examples of the annuity cash flow pattern.
Annuities may also be used to model series of cash flows that
fluctuate over time around some average value.
Here the average value would be the constant uniform cash
flow. This would be done if the fluctuations were unknown or
deemed to be unimportant for the problem.

Series present worth factor
The series present worth factor, denoted by (P/A,i,N), gives the
present amount, P, that is equivalent to an annuity with
disbursements or receipts in the amount, A, where the interest rate is
i and the number of periods is N.

Capital recovery factor (A/P)
The capital recovery factor, denoted by (A/P,i,N), gives the value,
A, of the equal periodic payments or receipts that are equivalent
to a present amount, P, when the interest rate is i and the
number of periods is N.
The capital recovery factor can be used to find out, for example, how much
money must be saved over N future periods to “recover” a capital investment
of P today.

Uniform Series Involving P/A and A/P
The cash flow diagrams are:
Standard Factor Notation
P = A(P/A,i,n) A = P(A/P,i,n)
Note: P is one period Ahead of first A value
The uniform series factors that involve P and A are derived as follows:
(1) Cash flow occurs in consecutive interest periods
(2) Cash flow amount is same in each interest period
2-29

Example: Uniform Series Involving P/A
A chemical engineer believes that by modifying the structure of a certain water
treatment polymer, his company would earn an extra $5000 per year. At an interest
rate of 10% per year, how much could the company afford to spend now to just
break even over a 5 year project period?
(A) $11,170 (B) 13,640 (C) $15,300 (D) $18,950
The cash flow diagram is as follows: Solution:
P = 5000(P/A,10%,5)
= 5000(3.7908)
= $18,954
Answer is (D)
2-31

Example 2.3
How much money should you be willing to pay now for a
guaranteed $600 per year for 9 years starting next year, at a
rate of return of 16% per year?

Example 2.4
The HBNA plant may generate a revenue base of $50 million per
year. The president of the Italian parent company Baleez may
have reason to be quite pleased with this projection for the
simple reason that over the 5-year planning horizon, the
expected revenue would total $250 million, which is $50 million
more than the initial investment. With money worth 10% per
year, address the following question from the president: Will the
initial investment be recovered over the 5-year horizon with the
time value of money considered? If so, by how much extra in
present worth funds? If not, what is the equivalent annual
revenue base required for the recovery plus the 10% return on
money? Use both tabulated factor values and spreadsheet
functions.

P = 100,000(P/A,12%,2)
= 100,000(1.6901)
= $169,010

A = 450,000(A/P,10%,3)
= 450,000(0.40211)
= $180,950

Sinking fund factor
The sinking fund factor, denoted by (A/F,i,N), gives the size, A,
of a repeated receipt or disbursement that is equivalent to a
future amount, F, if the interest rate is i and the number of
periods is N.
A sinking fund is an interest-bearing account into which regular
deposits are made in order to accumulate some amount.
The sinking fund factor is commonly used to determine how
much has to be set aside or saved per period to accumulate an
amount F at the end of N periods at an interest rate i.

Sinking fund factor
The amount F might be used, for example, to purchase new or
replacement equipment, to pay for renovations, or to cover
capacity expansion costs.
In more general terms, the sinking fund factor allows us to
convert a single future amount into a series of equal-sized
payments, made over N equally spaced intervals, with the use
of a given interest rate i.

Uniform series compound amount
factor
Uniform series compound amount factor, denoted by (F/A,i,N),
gives the future value, F, that is equivalent to a series of equal-
sized receipts or disbursements, A, when the interest rate is i
and the number of periods is N.

Uniform Series Involving F/A and A/F
The uniform series factors that involve F and A are derived as follows:
(1) Cash flow occurs in consecutive interest periods
(2) Last cash flow occurs in same period as F
Note: F takes place in the same period as last A
Cash flow diagrams are:
Standard Factor Notation
F = A(F/A,i,n) A = F(A/F,i,n)
2-41

Example: Uniform Series Involving F/A
An industrial engineer made a modification to a chip manufacturing
process that will save her company $10,000 per year. At an interest
rate of 8% per year, how much will the savings amount to in 7 years?
(A) $45,300 (B) $68,500 (C) $89,228 (D) $151,500
The cash flow diagram is:
F = ?
0 1 2 3
i = 8%
4 5 6 7
Solution:
F = 10,000(F/A,8%,7)
= 10,000(8.9228)
= $89,228
Answer is (C)
A = $10,000
2-44

Example 2.5
The president of Ford Motor Company
wants to know the equivalent future
worth of a $1 million capital investment
each year for 8 years, starting 1 year from
now. Ford capital earns at a rate of 14%
per year.

F = 100,000,000/30(F/A,10%,30)
= 3,333,333(164.4940)
= $548,313,333

F = 12,000(F/A,10%,30)
= 12,000(164.4940)
= $1,973,928

Example 2.6
Consider the HBNA case presented at the outset of this chapter,
in which a projected $200 million investment can generate $50
million per year in revenue for 5 years starting 1 year after start-
up. A 10% per year time value of money has been used
previously to determine P, F, and A values. Now the president
would like the answers to a couple of new questions about the
estimated annual revenues. Use tabulated values, factor
formulas, or spreadsheet functions to provide the answers.

(a) What is the equivalent future worth of the estimated
revenues after 5 years at 10% per year?
(b) Assume that, due to the economic downturn, the
president predicts that the corporation will earn only 4.5% per
year on its money, not the previously anticipated 10% per
year. What is the required amount of the annual revenue
series over the 5-year period to be economically equivalent to
the amount calculated in (a)?

Factor Values for Untabulated i or n
3 ways to find factor values for untabulated i or n values
Use formula
Use spreadsheet function with corresponding P, F, or A value set to 1
Linearly interpolate in interest tables
Formula or spreadsheet function is fast and accurate
Interpolation is only approximate
2-54

Example: Untabulated i
Determine the value for (F/P, 8.3%,10)
Formula: F = (1 + 0.083)10 =2.2197
Spreadsheet: = FV(8.3%,10,,1) = 2.2197
Interpolation: 8% 2.1589
8.3% ------ x
9% ------ 2.3674
x = 2.1589 + [(8.3 - 8.0)/(9.0 - 8.0)][2.3674 – 2.1589]
= 2.2215
Absolute Error = 2.2215 – 2.2197 = 0.0018
OK
2-56
OK
(Too high)

Example 2.7
Determine the P∕A factor value for i = 7.75% and n = 10 years, using the
three methods described previously.

Arithmetic Gradients
• Assume a manufacturing engineer predicts that the
cost of maintaining a robot will increase by $5000
per year until the machine is retired. The cash flow
series of maintenance costs involves a constant
gradient, which is $5000 per year.
• An arithmetic gradient series is a cash flow series
that either increases or decreases by a constant
amount each period. The amount of change is
called the gradient.

• Formulas previously developed for an “A” series
have year-end amounts of equal value.
• In the case of a gradient, each year-end cash flow
is different, so new formulas must be derived.
• First, assume that the cash flow at the end of year
1 is the base amount of the cash flow series and,
therefore, not part of the gradient series.
• This is convenient because in actual applications,
the base amount is usually significantly different in
size compared to the gradient.

• For example, if you purchase a used car with a 1-
year warranty, you might expect to pay the
gasoline and insurance costs during the first year
of operation. Assume these cost $2500; that is,
$2500 is the base amount.
• After the first year, you absorb the cost of repairs,
which can be expected to increase each year.
• If you estimate that total costs will increase by
$200 each year, the amount the second year is
$2700, the third $2900, and so on to year n, when
the total cost is 2500 + (n − 1)200.

Arithmetic gradients change by the same amount each period
The cash flow diagram for the PG
of an arithmetic gradient is: G starts between periods 1 and 2
(not between 0 and 1)
This is because cash flow in year 1 is
usually not equal to G and is handled
separately as a base amount
(shown on next slide)
Note that PG is located Two Periods
Ahead of the first change that is equal
to G
Standard factor notation is
PG = G(P/G,i,n)
2-64

Typical Arithmetic Gradient Cash Flow
PT = ?
0 1
i = 10%
2 3 4 5
400
450
500
550
600
PA = ? PG = ?
0 1
i = 10%
2 3 4 5
100
150
200
+
This diagram = this base amount plus this gradient
PA = 400(P/A,10%,5)
50
PG = 50(P/G,10%,5)
PT = PA + PG = 400(P/A,10%,5) +50(P/G,10%,5)
0 1
i = 10%
2 3 4 5
Amount 400 400 400 400 400
in year 1
is base
amount
Amount in year1
is base amount
2-65

Converting Arithmetic Gradient to A
0 1
i = 10%
2 3 4 5
G
2G
3G
4G
0 1
i = 10%
2 3 4 5
A = ?
Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n)
0 1 2 3 4 5
G
2G
3G
4G
General equation when base amount is involved is
A = base amount + G(A/G,i,n)
For decreasing gradients,
change plus sign to minus
A = base amount - G(A/G,i,n)
2-66

Example: Arithmetic Gradient
Solution:
PT = 400(P/A,12%,5) + 30(P/G,12%,5)
= 400(3.6048) + 30(6.3970)
= $1,633.83
Answer is (B)
The cash flow could also be converted
into an A value as follows:
A = 400 + 30(A/G,12%,5)
= 400 + 30(1.7746)
= $453.24
2-67

Example 2.8
A local university has initiated a logo-licensing program
with the clothier Holister, Inc. Estimated fees (revenues)
are $80,000 for the first year with uniform increases to a
total of $200,000 by the end of year 9. Determine the
gradient and construct a cash flow diagram that
identifies the base amount and the gradient series.

Example 2.9
Neighboring parishes in Louisiana have agreed to pool road tax
resources already designated for bridge refurbishment. At a
recent meeting, the engineers estimated that a total of $500,000
will be deposited at the end of next year into an account for the
repair of old and safety-questionable bridges throughout the
area. Further, they estimated that the deposits will increase by
$100,000 per year for only 9 years thereafter, then cease.
Determine the equivalent (a) present worth, and (b) annual
series amounts, if public funds earn at a rate of 5% per year.

Geometric Gradient Series Factors
It is common for annual revenues and annual costs such as
maintenance, operations, and labor to go up or down by a
constant percentage.
for example, +5% or −3% per year. This change occurs every
year on top of a starting amount in the first year of the project.
A geometric gradient series is a cash flow series that either
increases or decreases by a constant percentage each period.
The uniform change is called the rate of change.
The geometric gradient series may also be used to model
inflation or deflation, productivity improvement or degradation,
and growth or shrinkage of market size, as well as many other
phenomena.

Geometric Gradients
Geometric gradients change by the same percentage each period
0
1 2 3 n
A1
1
A (1+g)1
4
A (1+g)2
1
A (1+g)n-1
Pg =?
There are no tables for geometric factors
Use following equation for g ≠ i:
Pg = A1{1- [(1+g)/(1+i)]n}/(i-g)
where: A = cash flow in period 1
1
g = rate of increase
If g = i, Pg =A1n/(1+i)
Note: If g is negative, change signs in front of both g values
Cash flow diagram for present worth
of geometric gradient
1
Note: g starts between
periods 1 and 2
2-76

Example: Geometric Gradient
10
1070
1145
1838
Pg =?
1
0
1000
Solution:
Pg = 1000[1-(1+0.07/1+0.12)10]/(0.12-0.07)
= $7,333
Answer is (b)
g = 7%
i = 12%
2 3 4
To find A, multiply Pg by(A/P,12%,10)
2-78

Example 2.11
A coal-fired power plant has upgraded an emission
control valve. The modification costs only $8000 and
is expected to last 6 years with a $200 salvage value.
The maintenance cost is expected to be high at $1700
the first year, increasing by 11% per year thereafter.
Determine the equivalent present worth of the
modification and maintenance cost by hand and by
spreadsheet at 8% per year.

Example 2.12
Now let’s go back to the proposed HBNA steel plant. The
revenue series estimate of $50 million annually is quite
optimistic, especially since there are other steel fabricators
operating in the area. Therefore, it is important to be sensitive in
our analysis to possibly declining and increasing revenue series,
depending upon the longer-term success of the plant’s marketing,
quality, and reputation.
Assume that revenue may start at $50 million by the end of the
first year, but then decreases geometrically by 12% per year
through year 5.
Determine the present worth and future worth equivalents of all
revenues during this 5-year time frame at the same rate used
previously, that is, 10% per year.

Unknown Interest Rate i
Solution: Can use either the P/A or A/P factor. Using A/P:
60,000(A/P,i%,10) = 16,000
(A/P,i%,10) = 0.26667
From A/P column at n = 10 in the interest tables, i is between 22% and 24%
2-19
Answer is (d)
(Usually requires a trial and error solution or interpolation in interest tables)
Procedure: Set up equation with all symbols involved and solve for i

Example 2.13
If Laurel made a $30,000 investment in a friend’s business and
received $50,000 five years later, determine the rate of return.

Example 2.14
Pyramid Energy requires that for each of its offshore
wind power generators $5000 per year be placed into a
capital reserve fund to cover unexpected major rework
on field equipment. In one case, $5000 was deposited
for 15 years and covered a rework costing $100,000 in
year 15.
What rate of return did this practice provide to the
company? Solve by hand and spreadsheet.

Unknown Recovery Period n
Unknown recovery period problems involve solving for n,
given i and 2 other values (P, F, orA)
(Like interest rate problems, they usually require a trial & error solution or interpolation in interest tables)
Procedure: Set up equation with all symbols involved and solve for n
A contractor purchased equipment for $60,000 that provided income of $8,000
per year. At an interest rate of 10% per year, the length of time required to recover
the investment was closest to:
(a) 10 years (b) 12 years (c) 15 years (d) 18 years
2-89
Solution: Can use either the P/A or A/P factor. Using A/P:
60,000(A/P,10%,n) = 8,000
(A/P,10%,n) = 0.13333
From A/P column in i = 10% interest tables, n is between 14 and 15 years Answer is (c)

Example 2.15
From the introductory comments about the HBNA
plant, the annual revenue is estimated to be $50 million.
All analysis thus far has taken place at 10% per year;
however, the parent company has made it clear that its
other international plants are able to show a 20% per
year return on the initial investment.
Determine the number of years required to generate
10%, 15%, and 20% per year returns on the $200
million investment at the new site.

Summary of Important Points
In P/A and A/P factors, P is one period ahead of first A
In F/A and A/F factors, F is in same period as last A
To find untabulated factor values, best way is to use formula or spreadsheet
For arithmetic gradients, gradient G starts between periods 1 and 2
Arithmetic gradients have 2 parts, base amount (year 1) and gradient amount
For geometric gradients, gradient g starts been periods 1 and 2
In geometric gradient formula, A1 is amount in period 1
To find unknown i or n, set up equation involving all terms and solve for i or n
2-93

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